User anton fonarev - MathOverflow

Answer by Anton Fonarev for Normal bundle of an exceptional divisor

2013-06-07T08:38:13Z

As for the first question, this is written in many places. Searching MO, by the way, is a good idea: http://mathoverflow.net/questions/16788/blowing-up-1-curves-effective-and-ample-divisors

Generally, if $X$ is a smooth subscheme in $Y$ with normal bundle $N$, then $N_{\tilde{X}}\tilde{Y}=\mathcal{O}_N(-1)$.

As for the second one, this is obvious after you combine the comment above with functoriality of the blow-up.

Answer by Anton Fonarev for Representations of $GL(n)$ containing $S^kV$

2013-01-11T11:24:42Z

Regarding the question of pairs, here is the answer (hope, I'm not mistaken).

Let me first change the notation a little bit. Suppose, we are looking for irreps $V_1=\Sigma^{\lambda}V$ and $V_2=\Sigma^{\mu}V$ where $\lambda = (\lambda_1,\ldots,\lambda_n)$ and $\mu = (\mu_1,\ldots,\mu_n)$ resp., such that there exists an embedding $S^kV\to V_1^*\otimes V_2$.

It's immediate that such an embedding exists iff $V_2=\Sigma^{\mu}V$ is an irreducible factor in $V_1\otimes S^kV=\Sigma^{\lambda}V\otimes S^kV$. The Littlewood-Richardson rule gives us the answer. Firstly, $k=|\mu|-|\lambda|$, where $|\lambda| = \sum_i\lambda_i$. Secondly, one has the following inequalities:

$\mu_1\geq \lambda_1$,
$\lambda_1\geq \mu_2 \geq \lambda_2$,
$\ldots$
$\lambda_{n-1}\geq \mu_n \geq \lambda_n$.

Finally, if such an embedding exists, it's unique by the very same L-R rule.

UPD: Forgot to mention that ~~the main question still seems to be very hard as even~~ checking that $\dim V_1=\dim V_2$ is a huge problem.

UPD 2: Let me answer your main question (despite UPD). Let's forget about representation theory and do some linear algebra. What we have is a surjective map $V_1\otimes U\to V_2$, where $\dim V_1 = \dim V_2$. Let the corresponding bilinear map be $h:V_1\times U\to V_2$. We want to find such $u\in U$ that $h_u = h(-, u)$ is invertible, equivalently, of maximal rank.

Well, this seems to be quite obvious: let $r$ be the maximal rank of $h_u$ among all $u\in U$. Suppose $r<\dim V_2$. Let $u_0\in U$ be such that $\mathrm{rk}\ h_{u_0}=r$. Take any $v\in V_2\setminus\mathrm{Im}\ h_{u_0}$. Then there exists such $u_1\in U$ that $v\in \mathrm{Im}\ h_{u_1}$. Now it's easy to see that the rank of a general linear combination $\alpha h_{u_0} + \beta h_{u_1} = h_{\alpha u_0+\beta u_1}$ will be greater than $r$.

Answer by Anton Fonarev for Completion of a completion

2012-10-10T09:27:53Z

Let $A$ be a commutative ring, $I\subset A$ be an ideal. Denote by $\hat{A}$ the completion with respect to the $I$-adic filtration.

One can consider the kernel $\hat{I}_n$ of the natural projection $\hat{A}\to A/I^n$. The obvious observation is: $\hat{A}/\hat{I}_n=A/I^n$. Thus, $\hat{A}=\varprojlim_n\hat{A}/\hat{I}_n$.

Now, it' clear that $I^n\hat{A}\subseteq\hat{I}_n\subseteq(\hat{I}_1)^n$. In Noetherian case they are always equal. This gives an affirmative answer to your question. In general they can differ. However, I can not come up with an example immediately.

Answer by Anton Fonarev for system of local coefficients on X, locally constant sheaves and orientation sheaves

2012-07-06T10:57:29Z

These are purely topological notions and have nothing to do with algebraic geometry in particular.

Let $M$ for simplicity be a topological manifold of dimension $n$. Then the orientation sheaf $\mathcal{L}_{or}(M)$ is the sheafification of the presheaf $U\mapsto H_n(M,M-U;\mathbb{Z})$. It's always a locally constant sheaf with stalks equal to $\mathbb{Z}$. One immediately checks that $\mathcal{L}_{or}$ is trivial if and only if $M$ is orientable. This definition can be generalized.

As for the references, I'd suggest checking A.Dimca, Sheaves in Topology or B.Iversen, Cohomology of Sheaves.

Answer by Anton Fonarev for Extension of projective module

2012-07-04T10:10:28Z

Given a short exact sequence $0\to F_1 \to F \to F_2\to 0$, one has $pd(F)\leq \max \left( pd(F_1),pd(F_2) \right)$ with equality except when $pd(F_2)=pd(F_1)+1$. Suppose that $pd_B(M)<\infty$. Then $pd_B(N) = pd_B(M)$.

Now, $N$ is projective if and only if $pd_B(N)=0$. Therefore, one has to ask for $pd_B(M)=0$, which is the same as to say that $M$ is projective as a $B$-module.

Answer by Anton Fonarev for Good overview of singularity theory

2012-03-06T13:07:32Z

If you are looking for a more topological treatment of the subject, there is a two-volume Singularities of Differentiable Maps by Arnold, Varchenko and Gusein-Zade.

Answer by Anton Fonarev for Betti numbers from virtual Betti numbers of a cell decomposition

2011-10-24T13:02:35Z

Sure, it is true. Let's check once again the definition of virtual Betti numbers:

For $X$ smooth and compact $b_i(X)=\dim_{\mathbb{Q}} H^i(X,\mathbb{Q})$.
For any closed $Y\subset X$ one has $b_i(X) = b_i(Y)+b_i(X-Y)$.

Now, whenever you have an algebraic cell decomposition $F^1\subset\ldots\subset F^{n-1}\subset F^n=X$, by (2) you have $b_i(X)=b_i(F^n-F^{n-1})+\ldots+b_i(F^2-F^1)+b_i(F^1)$, where each $F^i-F^{i-1}$ is a disjoint union of some affine spaces. It is left to see that $b_i(\mathbb{A}^j)=b_i(\mathbb{P}^j)-b_i(\mathbb{P}^{j-1})=\delta^i_{2j}$.

Now combine with the first Proposition in [Fresse, 4.1] and get that whenever $X$ is projective $b_i(X)=\dim_{\mathbb{Q}}H^i(X,\mathbb{Q})$. This holds because $X$ compact implies $H^i_c(X,\mathbb{Q})=H^i(X,\mathbb{Q})$.

Ok, let's see why this Proposition holds in our particular situation. Take the first step of the stratification: $F^{n-1}\subset F^n=X$. Denote $U=F^n-F^{n-1}$. Then there is a long exact sequence associated to a closed subset: $\ldots\to H^i_c(U)\to H^i_c(X)\to H^i_c(F^{n-1})\to H^{i+1}_c(U)\to\ldots$. All the coefficients are $\mathbb{Q}$. Now $U$ is a disjoint union of affine cells of dimension equal to $\dim X$ and $\dim F^{n-1}<\dim X$ (you can always achieve this by a proper choice of stratification). Note that $F^{n-1}$ also admits a cell decomposition. So we may assume that the statement holds for $F^{n-1}$ by an inductive argument. In Particular, $H^i_c(F^{n-1})=0$ for all $i>2\dim F^{n-1}$, thus, for all $i > 2(\dim X-1)$. Combine with the long exact sequence and the fact that $\dim H^i_c(\mathbb{A}^n)=\delta^i_{2n}$. This finishes the proof.

The reference for the long exact sequence associated to a closed subset will be Iversen, Cohomology of sheaves.

However, as far as I understand, having an algebraic cell decomposition is somewhat strong. Note that in general virtual Betti numbers can be negative.

Answer by Anton Fonarev for Action of $SL_{n+1}$ on couples of linear spaces in $\mathbb{P}^n$.

2011-10-06T15:53:34Z

Sure it does. Denote your variety by $X$. Consider this action on the corresponding vector space $V$. Then your projective subspaces correspond to linear subspaces $V_1$ and $V_2$ of dimension $m+1$ and $n-m+1$ that intersect transversely. There is an obvious projection from the variety $Y$ of all frames in $V$ to $X$ (take the linear hull of the first $m+1$ and the last $n-m+1$ vectors). Now, up to a multiplication of all the vectors in the frame by a common scalar, $SL_{n+1}$ acts transitively on $Y$. Thus, it acts transitively on $X$.

Answer by Anton Fonarev for flatness in complex analytic geometry

2011-09-13T15:13:52Z

Well, I definitely know nothing on the subject. However, there is something called Hironaka division, which gives some way to check flatness explicitly. Check Section 4.2 of these notes.

Answer by Anton Fonarev for exact sequence of logarithmic differential sheaves associated to an effective Cartier divisor on a smooth variety

2011-09-08T23:08:05Z

Here is a way to define this sheaf algebraically over any field of characteristic zero. Let $\mathrm{T}_{X}$ denote the tangent sheaf on $X$. Choose a local equation $\phi_U$ for $D$ on $U$. Consider the following submodule:

$\mathrm{T}_X(-\log\phi_U)=\ ${$\partial\in\mathrm{Der}(\mathcal{O}_X(U))\mid \partial\phi_U\in (\phi_U)$}$\ \subset \mathrm{T}_X(U)$.

It is an easy exercise to check that this does not depend on the choice of equation and glues into a sheaf $\mathrm{T}_X(-\log D)$. Now take the dual $\Omega^1_X(\log D)=\mathrm{T}_X(-\log D)^*$.

The good thing about having normal crossings is that in this case $\Omega^1_X(\log D)$ becomes a locally free sheaf.

For your second question, look at Definition 2.5 in this paper by Dolgachev, arXiv:math/0508044

Answer by Anton Fonarev for How to introduce notions of flat, projective and free modules?

2010-11-18T23:56:10Z

Another student answer:

Considering 2b, it seems a bit strange that you want to mention derived functors and not to go into $\mathbf{Tor}$ and $\mathbf{Ext}$. The problem with derived functors is that first of all, categories usually have enough injective objects. Secondly, projectives are just a suitable collection of objects to use, but not always. However, the universal property is really important here. Lifting morphisms is something extremely natural to ask (which is the exactness of $\mathbf{Hom}$). This is also a way to mention that projective modules are direct summands of free (which fits perfectly into you sequence of presentation).

Speaking of flat modules, I'd avoid saying about "varying nicely". It's not very convincing the first time you see it (like telling a child that to have a baby one should kiss his partner). I'd also be nice to somehow sneak in the adjointness of $\otimes$ and $\mathbf{Hom}$.

Comment by Anton Fonarev

2013-06-07T21:41:23Z

By functoriality I mean that the blow up of $X$ is the strict transform of $X$ in the blow up of the ambient $\mathbb{A}^4$. See Sándor's answer for details.

Comment by Anton Fonarev

2013-01-11T13:34:37Z

@Klim Due to dualization, in my notation $V_1=S^{k/2}V^*$. In particular, $\lambda = (0,\ldots,0,-k/2)$.

Comment by Anton Fonarev

2013-01-11T10:59:40Z

@Dragos "invariable" should be a misprint for invertible.

Comment by Anton Fonarev

2013-01-11T10:57:29Z

@Aakumadula It's clearly written that Efim asks for $T(x)$ to be invertible. You can regard $T(x)$ as an element in $Hom(V_1^*,V_2)$.

Comment by Anton Fonarev

2012-11-26T22:41:25Z

This has already been around MO: mathoverflow.net/questions/44060

Comment by Anton Fonarev

2012-11-21T10:23:28Z

I'm sorry, but this is on the edge of being a research level question (as required in mathoverflow.net/faq).

Comment by Anton Fonarev

2012-08-02T12:04:03Z

@zatilokum It means that this is a constant sheaf.

Comment by Anton Fonarev

2012-07-06T11:00:38Z

Could someone retag, please? Say, some "sheaf-theory" and "at.algebraic-topology".

Comment by Anton Fonarev

2012-07-04T11:05:49Z

@Fernando, this obviously happens as $pd_{\mathbf{Z}/4}(\mathbf{Z}/2)=\infty$. I've already put this assumption that I had in mind.

Comment by Anton Fonarev

2012-07-04T10:57:14Z

@a-fortiori Sure, thanks for pointing this out.

Comment by Anton Fonarev

2012-07-04T10:17:16Z

It's somehow weird to have a question starting with "two noncommutative rings" tagged ac.commutative-algebra. Could someone change it to homological?

Comment by Anton Fonarev

2012-06-22T23:19:17Z

Your definition of an exceptional pair is very special. A usual one asks only for $Ext^\bullet(F, E)=0$. A pair is called strong if $Ext^i(E, F)=0$ for all $i>0$. Also when one defines mutations, $Hom$ is taken in the derived sense.

Comment by Anton Fonarev

2012-04-14T23:51:58Z

This is not an appropriate question for MO. I'm getting more and more concerned with the number of such questions arising in here.

Comment by Anton Fonarev

2012-04-08T15:31:58Z

I really like this derived category point of view!

Comment by Anton Fonarev

2012-03-28T21:58:18Z

By the way, if I remember correctly, this is the proof written in Ried's book.